Mass of 1 mole of nitrogen \(\left(N_{2}\right)=2 \times N=2 \times 14=28 \mathrm{~g}\)
Mass of 3 moles of hydrogen \(\left(\mathrm{H}_{2}\right)=3[2 \times \mathrm{H}]\)
=3[2 x 1]=3[2]
=6 g
Mass of 2 moles of ammonia \(\left(\mathrm{NH}_{3}\right)=2\left[\mathrm{NH}_{3}\right]\)
=2[1 x N + 3 x H] 
=2[1 x14 + 3 x 1] 
=2[14 + 3] 
=2[17] = 34 g
\(\left.\begin{array}{c}1 \text { mole of nitrogen }(28 \mathrm{~g}) \\ + \\ 3 \text { mole of hydrogen }(6 \mathrm{~g})\end{array}\right\}=\text {2 moles of ammonia} (34 \mathrm{~g})\)

The atomicity of an element can be derived using Avogadro's hypothesis. Let us consider the following equation

According to Avogadro's law, 1 volume of any gas occupy 'n' number of molecules.

if n = 1 then

(i) 1 molecule of hydrogen chloride gas is made up of \(\frac{1}{2}\) molecule of hydrogen and \(\frac{1}{2}\) molecule of chlorine
(ii) But 1 molecule of hydrogen chloride contains one atom of hydrogen and I atom of chlorine
(iii) Hence \(\frac{1}{2}\) molecule of hydrogen = 1 atom of hydrogen
(iv) Or 1 molecule of hydrogen = 2 atoms of hydrogen
(v) So the atomicity of hydrogen is 2, and molecular formula is H₂
Similarly,
\(\frac{1}{2}\) molecule of chlorine = 1 atom of chlorine 1molecule of chlorine = 2 atoms of chlorine So the atomicity of chlorine is 2, and its molecular formula is Cl₂

Applications of Avogadro's hypothesis
(i) It explains Gay-Lussac's law.
(ii) It helps in the determination of atomicity of gases.
(iii) Molecular formula of gases can be derived using Avogadro's law.
(iv) It determines the relation between molecular mass and vapour density.
(v) It helps to determine Gram molar volume of all gases (i.e. 22.4 lit at S.T.P)

We consider B-11 isotope presents x % in nature. So B-10, isotope presents (1-x) %.
Average atomic mass = mass of B- 11+ mass of B-10
10.804 = x \(\times\) 11 + (1 - x) \(\times\) 10 
10.804 = 11 x + 10 -10 x 
10.804 = x + 10 
x = (10.804 - 10) 
x = 0.804
So, % relative abundance of B-11 is 0.804 x 10 = 80.4 %
% relative abundance of B -10 is (1 - x) x 100
=(1 - 0.804) x 100 
= 0.196 x 100 
= 19.6 %
% relative abundance of B-10 and B-11 are 19.6 % and 80.4 % respectively.

\(\text { Molar mass of } \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}=\mathrm{Al} \times 2+3[\mathrm{~S} \times 1+\mathrm{O} \times 4]\)
\(27 \times 2+3[32 \times 1+16 \times 4]\)
= 54 + 3[32 + 64]
= 54 + 3 [96]
Molar Mass = 54 + 288 = 342
\(\text { Mass } \% \text { of an element }=\frac{\text { mass of that element in the compound }}{\text { molar mass of the compound }} \times100\)
\(\text { Mass } \% \text { oxygen in } \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}=\frac{192}{342} \times 100 \)
\(=0.5614 \times 100=56.14 \% \)
\(\text { Mass } \% \text { of oxygen in } \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \text { is } 56.14 \%\)

Gram molecular of CO₂ = 44g
Number of molecules present in 44g of CO₂
= 6.023 X 10²³
∴ Number of molecules present in 11g of CO₂
= \(\frac{6.023\times{10}^{23}}{44}\)x 11
= 1.53 x 10²³ molecules.

Number of moles = \(\frac{Mass}{Atomic \ mass}\)
∴ Mass = Number of moles x Atomic mass
= 0.5 x 16
= 8g.

Number of moles = \(\frac{Mass}{Molecular \ mass}\)
= \(\frac{2}{40}\) = 0.05

Number of moles = \(\frac{Mass}{Atomic \ mass}\)
= \(\frac{81}{27}\) = 3 moles

Atomic masses of C= 12, O= 16
Gram molecular mass of CO₂ = 1(C) + 2(O) = 1(12) + 2(16) = 12 + 32 = 44
Gram molecular mass of CO₂ = 44g.